Question

The half-life of a radioactive substance is the time required for half of a sample to undergo radioactive decay, or for the quantity to fall to half its original amount. Carbon 14 has a half-life of 5,730 years. Suppose given samples of carbon 14 weigh (fraction 5/8) of a pound and (fraction 7/8) of a pound. What was the total weight of the samples 11460 years ago
(show work please)

Answer 1

That's two half lives. 5/8 * 2 * 2 = 20/8 = 5/2 pounds7/8 * 2 * 2 = 28/8 = 7/2 pounds

Answer 2

Answer:

Amount of C-14 taken were 2.5 pounds and 3.5 pounds respectively.

Step-by-step explanation:

Radioactive decay is an exponential process represented by

$A_{t}=A_{0}e^{-kt}$

where $A_{t}$ = Amount of the radioactive element after t years

$A_{0}$ = Initial amount

k = Decay constant

t = time in years

Half life period of Carbon-14 is 5730 years.

$\frac{A_{0} }{2}=A_{0}e^{-5730k}$

$\frac{1}{2}=e^{-5730k}$

Now we take ln (Natural log) on both the sides

$ln(\frac{1}{2})=ln[e^{-5730k}]$

-ln(2) = -5730kln(e)

0.69315 = 5730k

$k=\frac{0.69315}{5730}$

$k=1.21\times 10^{-4}$

Now we have to calculate the weight of samples of C-14 taken for the remaining quantities $\frac{5}{8}$ and $\frac{7}{8}$ of a pound.

$\frac{5}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}$

$\frac{5}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}$

$\frac{5}{8}=A_{0}e^{(-1.3863)}$

$A_{0}=\frac{5}{8}\times e^{1.3863}$

$A_{0}=\frac{5}{8}\times 4$

$A_{0}=\frac{5}{2}$

$A_{0}=2.5$ pounds

Similarly for $\frac{7}{8}$ pounds

$\frac{7}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}$

$\frac{7}{8}=A_{0}e^{(-1.21\times 10^{-4}\times 11460)}$

$\frac{7}{8}=A_{0}e^{(-1.3863)}$

$A_{0}=\frac{7}{8}\times e^{(1.3863)}$

$A_{0}=\frac{7}{8}\times 4$

$A_{0}=\frac{7}{2}$

$A_{0}=3.5$ pounds